Q.1 / Membership determination to those who got IQ scores at least 2 SD of WAIS mean . Of a group of 200 ppl, how many are eligible for membership to this society ?
A. 1 B. 2 C. 3 D. 4 E. 5


Q.2 /B lymphocytes with 200V gene,20D gene,50J gene segment in heavy chain variable domain and 100 V gene ,10J gene in light chain variable domain. In absence of recombinational inaccuracies, how many distinct idiotypes can b produced?
A.1000 B.4000 C.1*104 D.2* 105 E.2*108
(please explain me how to calculate these problems)

Q1: 2sd above, 2.5% ppl r there..
so ans is E

Q2: (i m nt sure for answer)

i will help you with the first question....on the second one i have some ideea but not very much, so i don't wanna post something wrong

so....for the first question: before starting my explanation, you should MEMORIZE a few things....in a group like this one there are 3 characteristics: mean, median and mode; a SD means something 'away' from mean.....question is how much? away from mean by +/- 1 SD are 68% from the group, +/- 2 SD are 95% of the group and 3SD are 99.7% of the group(you must remember this numbers)

for your case; your group has 200 people; in order to be eligible for a membership, that specific person must have the score 2SD above mean which mean he/she should be in those above 95%(aka 5%).....the tricky part here is that 5% with a 2 SD deviation are 'compose' from 2.5% which are above 2 SD and 2.5% which are bellow 2 SD; this question is asking about those people who are above 2 SD; so, the answer to your question is 2.5% of 200 are having an IQ above 2 SD which means 5 people will have this score and this should be your answer(in the same matter 2.5% are 2 SD bellow mean which means 5 people will score 2SD bellow mean).

i hope this will help you....good luck!

Thanks for your kindly explanation !!
Please anyone explanation for question 2 !!!!!

Q.2 /B lymphocytes with 200V gene,20D gene,50J gene segment in heavy chain variable domain and 100 V gene ,10J gene in light chain variable domain. In absence of recombinational inaccuracies, how many distinct idiotypes can b produced?
A.1000 B.4000 C.1*10^4 D.2* 10^5 E.2*10^8
(please explain me how to calculate these problems)[/quote]

Hi.

The answer you want Is E. For these questions it is pretty much straight multiplication, since Both the VDJ segments of heavy chain, and the VJ segments of light chain add to the idiotype (the part that binds antigen). So all you need to do is multiple how many different heavy chains you can have ([V]x[D]x[J] or 200 x 20 x 50) which is 20000 by the number of different types of light chain you can have ([10]x[J] or 100 x 10) 1000. These numbers multiplied together gives you 200,000,000 or in scientific notation written 2x10^8.

I hope this clarifies things. And i agree with what everyone else said on the first question.

[quote=aagupta2]Q.2 /B lymphocytes with 200V gene,20D gene,50J gene segment in heavy chain variable domain and 100 V gene ,10J gene in light chain variable domain. In absence of recombinational inaccuracies, how many distinct idiotypes can b produced?
A.1000 B.4000 C.1*10^4 D.2* 10^5 E.2*10^8
(please explain me how to calculate these problems)[/quote]

Hi.

The answer you want Is E. For these questions it is pretty much straight multiplication, since Both the VDJ segments of heavy chain, and the VJ segments of light chain add to the idiotype (the part that binds antigen). So all you need to do is multiple how many different heavy chains you can have ([V]x[D]x[J] or 200 x 20 x 50) which is 20000 by the number of different types of light chain you can have ([10]x[J] or 100 x 10) 1000. These numbers multiplied together gives you 200,000,000 or in scientific notation written 2x10^8.

I hope this clarifies things. And i agree with what everyone else said on the first question.[/quote]

Thanks for your explanation ! i m confused of the numbers that are given .i read 10^8 as 108 . so I could not calculate . thanks again .